Integrand size = 24, antiderivative size = 162 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^3} \, dx=\frac {1}{2} a c (4 b c+5 a d) \sqrt {c+d x^2}+\frac {1}{6} a (4 b c+5 a d) \left (c+d x^2\right )^{3/2}+\frac {a (4 b c+5 a d) \left (c+d x^2\right )^{5/2}}{10 c}+\frac {b^2 \left (c+d x^2\right )^{7/2}}{7 d}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{2 c x^2}-\frac {1}{2} a c^{3/2} (4 b c+5 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \]
1/6*a*(5*a*d+4*b*c)*(d*x^2+c)^(3/2)+1/10*a*(5*a*d+4*b*c)*(d*x^2+c)^(5/2)/c +1/7*b^2*(d*x^2+c)^(7/2)/d-1/2*a^2*(d*x^2+c)^(7/2)/c/x^2-1/2*a*c^(3/2)*(5* a*d+4*b*c)*arctanh((d*x^2+c)^(1/2)/c^(1/2))+1/2*a*c*(5*a*d+4*b*c)*(d*x^2+c )^(1/2)
Time = 0.20 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.82 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^3} \, dx=\frac {\sqrt {c+d x^2} \left (30 b^2 x^2 \left (c+d x^2\right )^3+35 a^2 d \left (-3 c^2+14 c d x^2+2 d^2 x^4\right )+28 a b d x^2 \left (23 c^2+11 c d x^2+3 d^2 x^4\right )\right )}{210 d x^2}-\frac {1}{2} a c^{3/2} (4 b c+5 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right ) \]
(Sqrt[c + d*x^2]*(30*b^2*x^2*(c + d*x^2)^3 + 35*a^2*d*(-3*c^2 + 14*c*d*x^2 + 2*d^2*x^4) + 28*a*b*d*x^2*(23*c^2 + 11*c*d*x^2 + 3*d^2*x^4)))/(210*d*x^ 2) - (a*c^(3/2)*(4*b*c + 5*a*d)*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])/2
Time = 0.25 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.88, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {354, 100, 27, 90, 60, 60, 60, 73, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^3} \, dx\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^2 \left (d x^2+c\right )^{5/2}}{x^4}dx^2\) |
| \(\Big \downarrow \) 100 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int \frac {\left (2 b^2 c x^2+a (4 b c+5 a d)\right ) \left (d x^2+c\right )^{5/2}}{2 x^2}dx^2}{c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x^2}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int \frac {\left (2 b^2 c x^2+a (4 b c+5 a d)\right ) \left (d x^2+c\right )^{5/2}}{x^2}dx^2}{2 c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x^2}\right )\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {1}{2} \left (\frac {a (5 a d+4 b c) \int \frac {\left (d x^2+c\right )^{5/2}}{x^2}dx^2+\frac {4 b^2 c \left (c+d x^2\right )^{7/2}}{7 d}}{2 c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x^2}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \left (\frac {a (5 a d+4 b c) \left (c \int \frac {\left (d x^2+c\right )^{3/2}}{x^2}dx^2+\frac {2}{5} \left (c+d x^2\right )^{5/2}\right )+\frac {4 b^2 c \left (c+d x^2\right )^{7/2}}{7 d}}{2 c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x^2}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \left (\frac {a (5 a d+4 b c) \left (c \left (c \int \frac {\sqrt {d x^2+c}}{x^2}dx^2+\frac {2}{3} \left (c+d x^2\right )^{3/2}\right )+\frac {2}{5} \left (c+d x^2\right )^{5/2}\right )+\frac {4 b^2 c \left (c+d x^2\right )^{7/2}}{7 d}}{2 c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x^2}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \left (\frac {a (5 a d+4 b c) \left (c \left (c \left (c \int \frac {1}{x^2 \sqrt {d x^2+c}}dx^2+2 \sqrt {c+d x^2}\right )+\frac {2}{3} \left (c+d x^2\right )^{3/2}\right )+\frac {2}{5} \left (c+d x^2\right )^{5/2}\right )+\frac {4 b^2 c \left (c+d x^2\right )^{7/2}}{7 d}}{2 c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x^2}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{2} \left (\frac {a (5 a d+4 b c) \left (c \left (c \left (\frac {2 c \int \frac {1}{\frac {x^4}{d}-\frac {c}{d}}d\sqrt {d x^2+c}}{d}+2 \sqrt {c+d x^2}\right )+\frac {2}{3} \left (c+d x^2\right )^{3/2}\right )+\frac {2}{5} \left (c+d x^2\right )^{5/2}\right )+\frac {4 b^2 c \left (c+d x^2\right )^{7/2}}{7 d}}{2 c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x^2}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{2} \left (\frac {a (5 a d+4 b c) \left (c \left (c \left (2 \sqrt {c+d x^2}-2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )\right )+\frac {2}{3} \left (c+d x^2\right )^{3/2}\right )+\frac {2}{5} \left (c+d x^2\right )^{5/2}\right )+\frac {4 b^2 c \left (c+d x^2\right )^{7/2}}{7 d}}{2 c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x^2}\right )\) |
(-((a^2*(c + d*x^2)^(7/2))/(c*x^2)) + ((4*b^2*c*(c + d*x^2)^(7/2))/(7*d) + a*(4*b*c + 5*a*d)*((2*(c + d*x^2)^(5/2))/5 + c*((2*(c + d*x^2)^(3/2))/3 + c*(2*Sqrt[c + d*x^2] - 2*Sqrt[c]*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]]))))/(2* c))/2
3.7.31.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Time = 3.00 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.94
| method | result | size |
| pseudoelliptic | \(\frac {-\frac {15 x^{2} \left (a d +\frac {4 b c}{5}\right ) d \,c^{2} a \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right )}{2}+\left (7 \left (\frac {9}{49} b^{2} x^{4}+\frac {22}{35} a b \,x^{2}+a^{2}\right ) x^{2} d^{2} c^{\frac {3}{2}}-\frac {3 d \left (-\frac {6}{7} b^{2} x^{4}-\frac {92}{15} a b \,x^{2}+a^{2}\right ) c^{\frac {5}{2}}}{2}+x^{2} \left (\frac {3 b^{2} c^{\frac {7}{2}}}{7}+d^{3} x^{2} \sqrt {c}\, \left (\frac {3}{7} b^{2} x^{4}+\frac {6}{5} a b \,x^{2}+a^{2}\right )\right )\right ) \sqrt {d \,x^{2}+c}}{3 \sqrt {c}\, d \,x^{2}}\) | \(153\) |
| default | \(\frac {b^{2} \left (d \,x^{2}+c \right )^{\frac {7}{2}}}{7 d}+a^{2} \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}}}{2 c \,x^{2}}+\frac {5 d \left (\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}}}{5}+c \left (\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3}+c \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )\right )\right )}{2 c}\right )+2 a b \left (\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}}}{5}+c \left (\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3}+c \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )\right )\right )\) | \(183\) |
| risch | \(-\frac {c^{2} a^{2} \sqrt {d \,x^{2}+c}}{2 x^{2}}+\frac {b^{2} d^{2} x^{6} \sqrt {d \,x^{2}+c}}{7}+\frac {3 b^{2} d c \,x^{4} \sqrt {d \,x^{2}+c}}{7}+\frac {3 b^{2} c^{2} x^{2} \sqrt {d \,x^{2}+c}}{7}+\frac {b^{2} c^{3} \sqrt {d \,x^{2}+c}}{7 d}+\frac {2 x^{4} d^{2} \sqrt {d \,x^{2}+c}\, a b}{5}+\frac {22 c d \,x^{2} \sqrt {d \,x^{2}+c}\, a b}{15}+\frac {46 c^{2} \sqrt {d \,x^{2}+c}\, a b}{15}-\frac {5 c^{\frac {3}{2}} \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right ) a^{2} d}{2}-2 c^{\frac {5}{2}} \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right ) a b +\frac {x^{2} d^{2} \sqrt {d \,x^{2}+c}\, a^{2}}{3}+\frac {7 c d \sqrt {d \,x^{2}+c}\, a^{2}}{3}\) | \(252\) |
1/3*(-15/2*x^2*(a*d+4/5*b*c)*d*c^2*a*arctanh((d*x^2+c)^(1/2)/c^(1/2))+(7*( 9/49*b^2*x^4+22/35*a*b*x^2+a^2)*x^2*d^2*c^(3/2)-3/2*d*(-6/7*b^2*x^4-92/15* a*b*x^2+a^2)*c^(5/2)+x^2*(3/7*b^2*c^(7/2)+d^3*x^2*c^(1/2)*(3/7*b^2*x^4+6/5 *a*b*x^2+a^2)))*(d*x^2+c)^(1/2))/c^(1/2)/d/x^2
Time = 0.27 (sec) , antiderivative size = 349, normalized size of antiderivative = 2.15 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^3} \, dx=\left [\frac {105 \, {\left (4 \, a b c^{2} d + 5 \, a^{2} c d^{2}\right )} \sqrt {c} x^{2} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (30 \, b^{2} d^{3} x^{8} + 6 \, {\left (15 \, b^{2} c d^{2} + 14 \, a b d^{3}\right )} x^{6} - 105 \, a^{2} c^{2} d + 2 \, {\left (45 \, b^{2} c^{2} d + 154 \, a b c d^{2} + 35 \, a^{2} d^{3}\right )} x^{4} + 2 \, {\left (15 \, b^{2} c^{3} + 322 \, a b c^{2} d + 245 \, a^{2} c d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{420 \, d x^{2}}, \frac {105 \, {\left (4 \, a b c^{2} d + 5 \, a^{2} c d^{2}\right )} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (30 \, b^{2} d^{3} x^{8} + 6 \, {\left (15 \, b^{2} c d^{2} + 14 \, a b d^{3}\right )} x^{6} - 105 \, a^{2} c^{2} d + 2 \, {\left (45 \, b^{2} c^{2} d + 154 \, a b c d^{2} + 35 \, a^{2} d^{3}\right )} x^{4} + 2 \, {\left (15 \, b^{2} c^{3} + 322 \, a b c^{2} d + 245 \, a^{2} c d^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{210 \, d x^{2}}\right ] \]
[1/420*(105*(4*a*b*c^2*d + 5*a^2*c*d^2)*sqrt(c)*x^2*log(-(d*x^2 - 2*sqrt(d *x^2 + c)*sqrt(c) + 2*c)/x^2) + 2*(30*b^2*d^3*x^8 + 6*(15*b^2*c*d^2 + 14*a *b*d^3)*x^6 - 105*a^2*c^2*d + 2*(45*b^2*c^2*d + 154*a*b*c*d^2 + 35*a^2*d^3 )*x^4 + 2*(15*b^2*c^3 + 322*a*b*c^2*d + 245*a^2*c*d^2)*x^2)*sqrt(d*x^2 + c ))/(d*x^2), 1/210*(105*(4*a*b*c^2*d + 5*a^2*c*d^2)*sqrt(-c)*x^2*arctan(sqr t(-c)/sqrt(d*x^2 + c)) + (30*b^2*d^3*x^8 + 6*(15*b^2*c*d^2 + 14*a*b*d^3)*x ^6 - 105*a^2*c^2*d + 2*(45*b^2*c^2*d + 154*a*b*c*d^2 + 35*a^2*d^3)*x^4 + 2 *(15*b^2*c^3 + 322*a*b*c^2*d + 245*a^2*c*d^2)*x^2)*sqrt(d*x^2 + c))/(d*x^2 )]
Time = 24.12 (sec) , antiderivative size = 570, normalized size of antiderivative = 3.52 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^3} \, dx=- \frac {5 a^{2} c^{\frac {3}{2}} d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{2} - \frac {a^{2} c^{2} \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{2 x} + \frac {2 a^{2} c^{2} \sqrt {d}}{x \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {2 a^{2} c d^{\frac {3}{2}} x}{\sqrt {\frac {c}{d x^{2}} + 1}} + a^{2} d^{2} \left (\begin {cases} \frac {c \sqrt {c + d x^{2}}}{3 d} + \frac {x^{2} \sqrt {c + d x^{2}}}{3} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{2}}{2} & \text {otherwise} \end {cases}\right ) - 2 a b c^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )} + \frac {2 a b c^{3}}{\sqrt {d} x \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {2 a b c^{2} \sqrt {d} x}{\sqrt {\frac {c}{d x^{2}} + 1}} + 4 a b c d \left (\begin {cases} \frac {c \sqrt {c + d x^{2}}}{3 d} + \frac {x^{2} \sqrt {c + d x^{2}}}{3} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{2}}{2} & \text {otherwise} \end {cases}\right ) + 2 a b d^{2} \left (\begin {cases} - \frac {2 c^{2} \sqrt {c + d x^{2}}}{15 d^{2}} + \frac {c x^{2} \sqrt {c + d x^{2}}}{15 d} + \frac {x^{4} \sqrt {c + d x^{2}}}{5} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{4}}{4} & \text {otherwise} \end {cases}\right ) + b^{2} c^{2} \left (\begin {cases} \frac {c \sqrt {c + d x^{2}}}{3 d} + \frac {x^{2} \sqrt {c + d x^{2}}}{3} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{2}}{2} & \text {otherwise} \end {cases}\right ) + 2 b^{2} c d \left (\begin {cases} - \frac {2 c^{2} \sqrt {c + d x^{2}}}{15 d^{2}} + \frac {c x^{2} \sqrt {c + d x^{2}}}{15 d} + \frac {x^{4} \sqrt {c + d x^{2}}}{5} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{4}}{4} & \text {otherwise} \end {cases}\right ) + b^{2} d^{2} \left (\begin {cases} \frac {8 c^{3} \sqrt {c + d x^{2}}}{105 d^{3}} - \frac {4 c^{2} x^{2} \sqrt {c + d x^{2}}}{105 d^{2}} + \frac {c x^{4} \sqrt {c + d x^{2}}}{35 d} + \frac {x^{6} \sqrt {c + d x^{2}}}{7} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{6}}{6} & \text {otherwise} \end {cases}\right ) \]
-5*a**2*c**(3/2)*d*asinh(sqrt(c)/(sqrt(d)*x))/2 - a**2*c**2*sqrt(d)*sqrt(c /(d*x**2) + 1)/(2*x) + 2*a**2*c**2*sqrt(d)/(x*sqrt(c/(d*x**2) + 1)) + 2*a* *2*c*d**(3/2)*x/sqrt(c/(d*x**2) + 1) + a**2*d**2*Piecewise((c*sqrt(c + d*x **2)/(3*d) + x**2*sqrt(c + d*x**2)/3, Ne(d, 0)), (sqrt(c)*x**2/2, True)) - 2*a*b*c**(5/2)*asinh(sqrt(c)/(sqrt(d)*x)) + 2*a*b*c**3/(sqrt(d)*x*sqrt(c/ (d*x**2) + 1)) + 2*a*b*c**2*sqrt(d)*x/sqrt(c/(d*x**2) + 1) + 4*a*b*c*d*Pie cewise((c*sqrt(c + d*x**2)/(3*d) + x**2*sqrt(c + d*x**2)/3, Ne(d, 0)), (sq rt(c)*x**2/2, True)) + 2*a*b*d**2*Piecewise((-2*c**2*sqrt(c + d*x**2)/(15* d**2) + c*x**2*sqrt(c + d*x**2)/(15*d) + x**4*sqrt(c + d*x**2)/5, Ne(d, 0) ), (sqrt(c)*x**4/4, True)) + b**2*c**2*Piecewise((c*sqrt(c + d*x**2)/(3*d) + x**2*sqrt(c + d*x**2)/3, Ne(d, 0)), (sqrt(c)*x**2/2, True)) + 2*b**2*c* d*Piecewise((-2*c**2*sqrt(c + d*x**2)/(15*d**2) + c*x**2*sqrt(c + d*x**2)/ (15*d) + x**4*sqrt(c + d*x**2)/5, Ne(d, 0)), (sqrt(c)*x**4/4, True)) + b** 2*d**2*Piecewise((8*c**3*sqrt(c + d*x**2)/(105*d**3) - 4*c**2*x**2*sqrt(c + d*x**2)/(105*d**2) + c*x**4*sqrt(c + d*x**2)/(35*d) + x**6*sqrt(c + d*x* *2)/7, Ne(d, 0)), (sqrt(c)*x**6/6, True))
Time = 0.21 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.05 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^3} \, dx=-2 \, a b c^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) - \frac {5}{2} \, a^{2} c^{\frac {3}{2}} d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) + \frac {2}{5} \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b + \frac {2}{3} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c + 2 \, \sqrt {d x^{2} + c} a b c^{2} + \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2}}{7 \, d} + \frac {5}{6} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d + \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} d}{2 \, c} + \frac {5}{2} \, \sqrt {d x^{2} + c} a^{2} c d - \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} a^{2}}{2 \, c x^{2}} \]
-2*a*b*c^(5/2)*arcsinh(c/(sqrt(c*d)*abs(x))) - 5/2*a^2*c^(3/2)*d*arcsinh(c /(sqrt(c*d)*abs(x))) + 2/5*(d*x^2 + c)^(5/2)*a*b + 2/3*(d*x^2 + c)^(3/2)*a *b*c + 2*sqrt(d*x^2 + c)*a*b*c^2 + 1/7*(d*x^2 + c)^(7/2)*b^2/d + 5/6*(d*x^ 2 + c)^(3/2)*a^2*d + 1/2*(d*x^2 + c)^(5/2)*a^2*d/c + 5/2*sqrt(d*x^2 + c)*a ^2*c*d - 1/2*(d*x^2 + c)^(7/2)*a^2/(c*x^2)
Time = 0.30 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^3} \, dx=\frac {30 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} + 84 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b d + 140 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c d + 420 \, \sqrt {d x^{2} + c} a b c^{2} d + 70 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{2} + 420 \, \sqrt {d x^{2} + c} a^{2} c d^{2} - \frac {105 \, \sqrt {d x^{2} + c} a^{2} c^{2} d}{x^{2}} + \frac {105 \, {\left (4 \, a b c^{3} d + 5 \, a^{2} c^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}}}{210 \, d} \]
1/210*(30*(d*x^2 + c)^(7/2)*b^2 + 84*(d*x^2 + c)^(5/2)*a*b*d + 140*(d*x^2 + c)^(3/2)*a*b*c*d + 420*sqrt(d*x^2 + c)*a*b*c^2*d + 70*(d*x^2 + c)^(3/2)* a^2*d^2 + 420*sqrt(d*x^2 + c)*a^2*c*d^2 - 105*sqrt(d*x^2 + c)*a^2*c^2*d/x^ 2 + 105*(4*a*b*c^3*d + 5*a^2*c^2*d^2)*arctan(sqrt(d*x^2 + c)/sqrt(-c))/sqr t(-c))/d
Time = 5.83 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.69 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^3} \, dx=\sqrt {d\,x^2+c}\,\left (c^2\,\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d}-\frac {2\,b^2\,c}{d}\right )-2\,c\,\left (2\,c\,\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d}-\frac {2\,b^2\,c}{d}\right )-\frac {{\left (a\,d-b\,c\right )}^2}{d}+\frac {b^2\,c^2}{d}\right )\right )-\left (\frac {2\,b^2\,c-2\,a\,b\,d}{5\,d}-\frac {2\,b^2\,c}{5\,d}\right )\,{\left (d\,x^2+c\right )}^{5/2}-{\left (d\,x^2+c\right )}^{3/2}\,\left (\frac {2\,c\,\left (\frac {2\,b^2\,c-2\,a\,b\,d}{d}-\frac {2\,b^2\,c}{d}\right )}{3}-\frac {{\left (a\,d-b\,c\right )}^2}{3\,d}+\frac {b^2\,c^2}{3\,d}\right )+\frac {b^2\,{\left (d\,x^2+c\right )}^{7/2}}{7\,d}-\frac {a^2\,c^2\,\sqrt {d\,x^2+c}}{2\,x^2}+\frac {a\,c^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {d\,x^2+c}\,1{}\mathrm {i}}{\sqrt {c}}\right )\,\left (5\,a\,d+4\,b\,c\right )\,1{}\mathrm {i}}{2} \]
(c + d*x^2)^(1/2)*(c^2*((2*b^2*c - 2*a*b*d)/d - (2*b^2*c)/d) - 2*c*(2*c*(( 2*b^2*c - 2*a*b*d)/d - (2*b^2*c)/d) - (a*d - b*c)^2/d + (b^2*c^2)/d)) - (( 2*b^2*c - 2*a*b*d)/(5*d) - (2*b^2*c)/(5*d))*(c + d*x^2)^(5/2) - (c + d*x^2 )^(3/2)*((2*c*((2*b^2*c - 2*a*b*d)/d - (2*b^2*c)/d))/3 - (a*d - b*c)^2/(3* d) + (b^2*c^2)/(3*d)) + (b^2*(c + d*x^2)^(7/2))/(7*d) + (a*c^(3/2)*atan((( c + d*x^2)^(1/2)*1i)/c^(1/2))*(5*a*d + 4*b*c)*1i)/2 - (a^2*c^2*(c + d*x^2) ^(1/2))/(2*x^2)